The problem is not the filename. The problem is that you have prototyped display(int,int) but not defined it. That is why you are getting a...
Just thought I'd borrow something from IOCCC to show you people what a difficult C program really looks like. These piddly examples to draw stars...
How far have you got and where are you stuck? Do you understand the requirements?
If you only have one process that writes to the table then you could do something like SELECT MAX() on a numeric column, add 1, then make that the...
A segfault means your program is writing to some memory it shouldn't be writing to (typically a NULL pointer).
"Why do you want to know" is not a difficult question to understand. But I can't answer your original question until I know, er, why you want to...
Whatever result you see is undefined because you are using modifiers more than once on the same variable in the same statement.
This is done in a testbed so my main function calls test56(). You might want to rename test56 to main. I couldn't be bothered typing stuff in so...
OK it's probably easiest if I give you an example. class housing { private: int REG_NO; char NAME[10]; char TYPE; float COST; public: void...
housing* h[5]; defines h as an array of 5 pointers to housing objects. You don't define the memory for those objects anywhere; these are only...
Why do you want to know? They are loaded with main in the TEXT part of the executable. You will need to learn assembly language programming to...
Don't forget that x++ means POSTincrement and ++x means PREincrement. That means before and after, and is usually before or after some execution...
This is down to the way scanf works. When you enter a number you press 1, 0, Enter. scanf reads the 1 and 0, then sees that the next character...
char is a single character, not a string, and %c is the format string for reading a single character. Try defining x as an array of char instead...
It's casting b, which is an integer, to a char*, then casting a, which is an array but taken as a char* in this context, to an integer, then using...
Oh I see. You're thinking that: printf("A","B","C"); will print ABC, right? It won't. The first parameter to printf is called a format...
What do you think the following code will display? #include <stdio.h> char*s="AAA %c %s %c BBB"; main() { printf(s,'\"',s,'\"'); }
Sorry, don't understand. What output are you expecting and why?
My favourite debugging technique is to get in there with a debugger, or failing that to sprinkle printf statements liberally across the code to...
It's all in the printf format string. The first parameter to printf is s, the string itself. So this is equivalent to: #include <stdio.h>...
Separate names with a comma.
